Problem: You have found the following ages (in years) of all 6 sloths at your local zoo: $ 17,\enspace 10,\enspace 1,\enspace 17,\enspace 11,\enspace 14$ What is the average age of the sloths at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we have data for all 6 sloths at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{17 + 10 + 1 + 17 + 11 + 14}{{6}} = {11.7\text{ years old}} $ Find the squared deviations from the mean for each sloth. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $17$ years $5.3$ years $28.09$ years $^2$ $10$ years $-1.7$ years $2.89$ years $^2$ $1$ year $-10.7$ years $114.49$ years $^2$ $17$ years $5.3$ years $28.09$ years $^2$ $11$ years $-0.7$ years $0.49$ years $^2$ $14$ years $2.3$ years $5.29$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{28.09} + {2.89} + {114.49} + {28.09} + {0.49} + {5.29}} {{6}} $ $ {\sigma^2} = \dfrac{{179.34}}{{6}} = {29.89\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{29.89\text{ years}^2}} = {5.5\text{ years}} $ The average sloth at the zoo is 11.7 years old. There is a standard deviation of 5.5 years.